Question: Find the limit as $x$ approaches negative infinity. $\lim_{x\to-\infty}\dfrac{\sqrt{4x^4-x}}{2x^2+3}=$
Solution: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the denominator is $x^2$, let's divide by $x^2$. In the numerator, let's divide by $\sqrt{x^4}$, since for any value, $x^2=\sqrt{x^4}$. $\begin{aligned} &\phantom{=}\lim_{x\to-\infty}\dfrac{\sqrt{4x^4-x}}{2x^2+3} \\\\ &=\lim_{x\to-\infty}\dfrac{\dfrac{\sqrt{4x^4-x}}{\sqrt{x^4}}}{\dfrac{2x^2+3}{x^2}} \gray{\text{Divide sides by }x^2=\sqrt{x^4}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to-\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to-\infty}\dfrac{\sqrt{\dfrac{4\cancel{x^4}}{\cancel{x^4}}-\dfrac{1\cancel x}{\cancel x\cdot x^3}}}{\dfrac{2\cancel{x^2}}{\cancel{x^2}}+\dfrac{3}{x^2}} \\\\ &=\lim_{x\to-\infty}\dfrac{\sqrt{4-\dfrac{1}{x^3}}}{2+\dfrac{3}{x^2}} \\\\ &=\lim_{x\to-\infty}\dfrac{\sqrt{4-0}}{2+0} \gray{\lim_{x\to-\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{\sqrt{4}}{2} \\\\ &=\dfrac{2}{2} \\\\ &=1 \end{aligned}$ In conclusion, $\lim_{x\to -\infty}\dfrac{\sqrt{4x^4-x}}{2x^2+3}=1$.